Question

The coefficient of volume expansion of a liquid is $49 \times {10^{ - 5}}{K^{ - 1}}$. Calculate the fractional change in its density when the temperature is raised by $30^\circ C$
(A) $7.5 \times {10^{ - 2}}$
(B) $3.0 \times {10^{ - 2}}$
(C) $1.5 \times {10^{ - 2}}$
(D) $1.1 \times {10^{ - 2}}$

Answer 1

From the coefficient of volume expansion and the temperature change, the final volume can be calculated by the formula, $V = {V_o}\left( {1 + \gamma \Delta T} \right)$. Since the density is inversely proportional to the volume, we can find the value of the fraction of initial by final density in terms of the volume. And then using that we can find the fractional change in density by $\dfrac{{{\rho _o} - \rho }}{{{\rho _o}}}$ .
Formula used: In the solution to this question, we will be using the following formula,
$\Rightarrow V = {V_o}\left( {1 + \gamma \Delta T} \right)$
where $V$ is the final volume and ${V_o}$ is the initial volume.
$\gamma$ is the coefficient of volume expansion and $\Delta T$ is the temperature change.
And the fractional change in density is given by $\dfrac{{{\rho _o} - \rho }}{{{\rho _o}}}$
where ${\rho _o}$ is the initial and $\rho$ is the final density.

Complete step by step answer:
In the question, we are given the coefficient of volume expansion $\gamma = 49 \times {10^{ - 5}}{K^{ - 1}}$. Now for a change in temperature, we can find the final volume of the liquid by the formula,
$\Rightarrow V = {V_o}\left( {1 + \gamma \Delta T} \right)$
We are provided the change of temperature as, $\Delta T = 30^\circ C$.
So the final volume of the liquid by substituting the values is,
$\Rightarrow V = {V_o}\left( {1 + 49 \times {{10}^{ - 5}} \times 30} \right)$
By bringing ${V_o}$ to the L.H.S of the equation we get,
$\Rightarrow \dfrac{V}{{{V_o}}} = 1 + 1.47 \times {10^{ - 2}}$
On doing the calculation, we get
$\Rightarrow \dfrac{V}{{{V_o}}} = 1.0147$
Now according to the basic definition of density, it is given by mass per unit volume. So, $\Rightarrow \rho = \dfrac{m}{V}$
Hence density is inversely proportional to the volume when the mass is constant.
$\Rightarrow \rho \propto \dfrac{1}{V}$
$\Rightarrow \rho V = {\text{constant}}$ by bringing the volume to the L.H.S
therefore since the product of volume and density is constant, we can write,
$\Rightarrow \rho V = {\rho _o}{V_o}$
By taking similar terms on one side,
$\Rightarrow \dfrac{V}{{{V_o}}} = \dfrac{{{\rho _o}}}{\rho }$
So from the previously derived value, we can write,
$\Rightarrow \dfrac{{{\rho _o}}}{\rho } = 1.0147$
Taking the reciprocal,
$\Rightarrow \dfrac{\rho }{{{\rho _o}}} = \dfrac{1}{{1.0147}} = 0.98558$
Now the fractional change in density is found out by the formula,
the fractional change in density $= \dfrac{{{\rho _o} - \rho }}{{{\rho _o}}}$
we can write this as,
$\Rightarrow 1 - \dfrac{\rho }{{{\rho _o}}}$
Now substituting the value of $\dfrac{\rho }{{{\rho _o}}}$ from the above equation we get,
$\Rightarrow 1 - 0.98558$
By doing the calculation we get the fractional change in density as,
$\Rightarrow 0.0145 = 1.45 \times {10^{ - 2}}$
This value can be rounded off to, $1.5 \times {10^{ - 2}}$
So the fraction change in density is $1.5 \times {10^{ - 2}}$
Therefore the correct answer is option (C).

Note:
When thermal energy is provided to liquids, they expand or contract according to the change in its temperature. The amount of expansion or contraction of the liquid with temperature depends on its coefficient of volume expansion. This coefficient has a different value for different liquids.